To prove it take everywhere the middles : {H, N, K, M, G, L, J} respectively of {EA, ED, DC,

CF, FB, BD, AC}. The quadrilaterals q

[1] Lines {EM, FM} which are parallel to {AD, BC} from {E,F} respectively, intersect at a point M on line KI joining the middles of the parallel sides of the trapezium.

[2] Let {E',F'} be the projections of {E, F} on sides {AD, BC} respectively. The ratio EE'/FF' = BC/AD, thus it is constant and independent of the particular direction of the parallels {e,e',f,f'}.

[2] Follows immediately from the preceding paragraph. [1] follows from [2]. In fact, extend DC until to intersect the parallels at {D',C'}. Then DD'=EE'/sin(D) and CC'=FF'/sin(C). Dividing the terms and using the parallelity from which sin(C)/sin(D)=sin(B)/sin(A)=AD/BC, follows that DD'=CC'. Thus {D',C'} are symmetric with respect to K and consequently M is on LK.

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